Question

2. Complete the square to determine the vertex and axis of symmetry of y=−x2+8x+7
. 3. Find the zeros of the quadratic equation y=6x2+13x+5
. You must use one of the following methods: factoring, square roots, complete the square, or quadratic formula.

Answer 1

Let's tackle each part of your question step by step.

### Part 2: Completing the Square

To find the vertex and axis of symmetry for the quadratic function:

\[ y = -x^2 + 8x + 7 \]

1. First, we want to rearrange the quadratic into the standard form \( y = a(x - h)^2 + k \). We will complete the square.

2. Factor out the coefficient of \( x^2 \) from the quadratic term and the linear term:

\[ y = -(x^2 - 8x) + 7 \]

3. Next, we complete the square inside the parentheses. To do this, take half of the coefficient of \( x \) (which is -8), square it, and add and subtract it inside the parentheses. Half of -8 is -4, and squaring it gives 16.

\[ y = -\left(x^2 - 8x + 16 - 16\right) + 7 \]

4. Rewrite it:

\[ y = -\left((x - 4)^2 - 16\right) + 7 \]

5. Distribute the negative sign:

\[ y = -(x - 4)^2 + 16 + 7 \]
\[ y = -(x - 4)^2 + 23 \]

Now we have it in vertex form.

### Vertex and Axis of Symmetry

From the expression \( y = -(x - 4)^2 + 23 \):

- The vertex \((h, k)\) is \((4, 23)\).
- The axis of symmetry is the line \( x = h \), which in this case is:

\[ x = 4 \]

### Part 3: Finding the Zeros of the Quadratic Equation

To find the zeros of the quadratic equation:

\[ y = 6x^2 + 13x + 5 \]

We can use the **quadratic formula**, which is given by:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

In our equation, \( a = 6 \), \( b = 13 \), and \( c = 5 \).

1. Calculate the discriminant \( b^2 - 4ac \):

\[
b^2 - 4ac = 13^2 - 4(6)(5) = 169 - 120 = 49
\]

2. Now we apply the quadratic formula:

\[
x = \frac{-13 \pm \sqrt{49}}{2 \times 6}
\]

3. Since \( \sqrt{49} = 7 \), we find:

\[
x = \frac{-13 \pm 7}{12}
\]

4. This gives us two solutions:

- For the positive root:

\[
x = \frac{-13 + 7}{12} = \frac{-6}{12} = -\frac{1}{2}
\]

- For the negative root:

\[
x = \frac{-13 - 7}{12} = \frac{-20}{12} = -\frac{5}{3}
\]

### Final Answers

1. Vertex of \( y = -x^2 + 8x + 7 \): \((4, 23)\)
Axis of symmetry: \( x = 4 \)

2. Zeros of \( y = 6x^2 + 13x + 5 \):
\[ x = -\frac{1}{2} \quad \text{and} \quad x = -\frac{5}{3} \]